# Show that `101^50 > 99^50 + 100^50`

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

`101^50`
`= (100 + 1)^50`
`= 100^50 + (^50C_1)*100^49 + (^50C_2)*100^48 + (^50C_3)*100^47 + ... + (^50C_49)*100^1 + 1`

`99^50 = (100 - 1)^50 = 100^50 - (^50C_1)*100^49 + (^50C_2)*100^48 - (^50C_3)*100^47 + ... - (^50C_49)*100^1 + 1`

Therefore, `(101^50 - 99^50)`
`= 2 [(^50C_1)*100^49 + (^50C_3)*100^47 + (^50C_5)*100^45 + ... + (^50C_49)*100^1]`
`= 2 [(50*100^49) + (19600*100^47) + .... ]`

Computing the first two terms, and ignoring the remaining positive terms we get,
`(101^50 - 99^50) = 2 [(50*100^49) + (1.96*100^49) + .... ]`
`= 2 [(50+1.96)*100^49 + ... ]`
`= 2 [51.96*100^49 + ... ]`
`= 103.92*100^49 + ...`
`= 1.0392*100^50 + ...`
`gt 100^50`

Thus,
`101^50 - 99^50 gt 100^50`
``
`rArr 101^50 gt 99^50 + 100^50`

juniorsilvamath | Student, Undergraduate | (Level 1) Honors

Posted on

Be a = 100^50 e b = 100^50 + 99^50, then,

a = (100+1)^50 = C(0,50)100^50+C(1,50)100^49+...+C(50,50)

b = 100^50+(100-1)^50 = C(0,50)100^50 - C(1,50)100^49+C(2,50)100^48-...-C(49,50)100+C(50,50)+100^50

Soon,

a-b=2C(1,50)100^49+2C(3,50)100^47+...+2C(49,50)100-100^50

= 2C(1,50)100^49 + ... + 2C(49,50)100> 0

Therefore, a > b

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